Problem: Simplify the following expression: $y = \dfrac{-8x^2+45x- 25}{x - 5}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-8)}{(-25)} &=& 200 \\ {a} + {b} &=& &=& {45} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $200$ and add them together. The factors that add up to ${45}$ will be your ${a}$ and ${b}$ When ${a}$ is ${5}$ and ${b}$ is ${40}$ $ \begin{eqnarray} {ab} &=& ({5})({40}) &=& 200 \\ {a} + {b} &=& {5} + {40} &=& 45 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-8}x^2 +{5}x) + ({40}x {-25}) $ Factor out the common factors: $ x(-8x + 5) - 5(-8x + 5)$ Now factor out $(-8x + 5)$ $ (-8x + 5)(x - 5)$ The original expression can therefore be written: $ \dfrac{(-8x + 5)(x - 5)}{x - 5}$ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ This leaves us with $-8x + 5; x \neq 5$.